【发布时间】:2020-05-11 00:25:03
【问题描述】:
处理程序有一个类继承器,我通过它控制图形按钮。 为了搜索这个按钮,我在设计器中传递了一个激活链接。但我想通过那里的观点。如何从活动中获取视图?
public class FlActivity : AppCompatActivity
{
protected override void OnCreate(Bundle savedInstanceState)
{
base.OnCreate(savedInstanceState);
SetContentView(Resource.Layout.activity_fl);
toolbar = FindViewById<Android.Support.V7.Widget.Toolbar>(Resource.Id.toolbar);
SetSupportActionBar(toolbar);
SupportActionBar.SetDisplayHomeAsUpEnabled(true);
tabLayout = FindViewById<TabLayout>(Resource.Id.tabLayout);
viewPager = FindViewById<ViewPager>(Resource.Id.viewPager);
fpAdapter = new FpAdapter(SupportFragmentManager, null);
uiHandler = new UiHandler(this, MainLooper);
}
}
public class UiHandler : Handler
{
private Activity activity { get; }
public UiHandler(Activity a, Looper loader) : base(loader)
{
activity = a;
}
public override void HandleMessage(Message msg)
{
activity.FindViewById<ImageButton>(Resource.Id.imageButton1).SetImageResource(msg.Data.GetInt("ID", Resource.Drawable.bt_off));
}
}
如果我将private Activity activity { get; } 更改为private View view { get; },如何在创建处理程序实例时从主活动转移视图。在创建 nadler 对象时用什么替换 this?
【问题讨论】:
标签: c# android xamarin android-activity view